Pairwise Rational Points on a Parabola

TL;DR

This paper explores the longstanding open problem of determining the maximum number of points on the parabola y=x^2 with pairwise rational distances, and demonstrates the existence of infinitely many such configurations through a constructive algorithm.

Contribution

It introduces a method to construct infinitely many families of points on the parabola with pairwise rational distances, advancing understanding of this open problem.

Findings

Existence of infinitely many configurations of points with rational pairwise distances.

A constructive algorithm for generating such point families.

Demonstration that arbitrarily large sets of such points can be constructed.

Abstract

This paper presents a solution to the following open problem in Number Theory and Geometry: How many points can you find on the (half) parabola $y=x^2$, $x>0$, so that the distance between any pair of them is rational? This problem sounds like a geometry problem, but it is likely to require techniques in number theory. That's because, to determine if the distance between (a,b) and (s,t) is rational, you need to test whether $(a-s)^2$ + $(b-t)^2$ is the square of a rational number...and such questions typically fall into the realm of number theory. Of course, since this is an open problem, no-one can claim to know just what field the problem lies in, since no-one knows the solution. I believe it is not even known if there are more than 5 points on the parabola which satisfy the condition. It is certainly not known if there is an infinite family with pairwise rational distances. We can quickly see that there are three such points in the following way: Pick two rational numbers, say 1/5 and 3/5, and let those be the distances between the pairs of points AB and BC. If you fix those distances and slide the points up the parabola, the distance AC will gradually increase, bounded above by 4/5. Since the rational numbers are dense in the reals, there will be many placements of the points so that AC is a rational distance. It is not too hard to show that if you have N points on the parabola with rational distances between them, then you can find N points on the parabola with integer distances between them. This problem is publicly posted on the DIMACS mathematical research website in order to search for solutions. This paper demonstrates that an infinite number of families each containing an arbitrarily large number of such pairwise rational points on the parabola can be found. A constructive algorithm for doing so is outlined in the paper.

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Pairwise Rational Points on a Parabola

Kyle Bomeisl

Abstract

This paper presents a solution to an open problem in Number Theory and Geometry. The problem is of public interest and is being actively researched by professional mathematicians. The original problem is publicly posted on the DIMACS mathematical research website by Professor Dean:

The problems is this: How many points can you find on the (half) parabola $y=x^{2}$, $x>0$, so that the distance between any pair of them is rational? This problem sounds like a geometry problem, but it is likely to require techniques in number theory. That’s because, to determine if the distance between (a,b) and (s,t) is rational, you need to test whether $(a-s)^{2}$ + $(b-t)^{2}$ is the square of a rational number…and such questions typically fall into the realm of number theory. Of course, since this is an open problem, no-one can claim to know just what field the problem lies in, since no-one knows the solution. I believe it is not even known if there are more than 5 points on the parabola which satisfy the condition. It is certainly not known if there is an infinite family with pairwise rational distances. We can quickly see that there are three such points in the following way: Pick two rational numbers, say 1/5 and 3/5, and let those be the distances between the pairs of points AB and BC. If you fix those distances and slide the points up the parabola, the distance AC will gradually increase, bounded above by 4/5. Since the rational numbers are dense in the reals, there will be many placements of the points so that AC is a rational distance. It is not too hard to show that if you have N points on the parabola with rational distances between them, then you can find N points on the parabola with integer distances between them. [1]

In this paper, it is demonstrated that there are an infinite number of families each containing an arbitrarily large number of pairwise rational points.

Introduction

In the first section, the distance between two given points is written down in terms of the x and y coordinates of each points. The coordinates used are in terms of the x and y axes that define the constraint parabola ($y=x^{2}$). Then the expression for the distance between two given points on the constraint parabola is factored into the form $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$. The game is now to constrain $(x_{2}-x_{1})$ and $\sqrt{1+(x_{2}+x_{1})^{2}}$ separately in order to force the overall expression $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$ to be rational. There are a limited number of classes of numbers that can be multiplied by each other and produce a rational product. The first class of numbers we test is a rational number multiplied by another rational number. Thus, $(x_{2}-x_{1})$ and $\sqrt{1+(x_{2}+x_{1})^{2}}$ must both be rational numbers. When we look for more than a few points whose x coordinates satisfy these requirements, the problem becomes fruitless and a proper solution (in the form of a family of points that are pairwise rational) is not found. This eliminates the product of two rational numbers as the form of the solutions. A rational number multiplied by an irrational number is always irrational, so it is necessary to search for solutions where $(x_{2}-x_{1})$ and $\sqrt{1+(x_{2}+x_{1})^{2}}$ are irrational. The first such form of two such irrational numbers that is guaranteed to produce a rational number when multiplied is $Q_{1}\sqrt{Q_{2}}$ and $Q_{3}\sqrt{Q_{2}}$. However, this again fails to produce solutions. The final set of constraints tested, which does bear fruit, is setting $(x_{2}-x_{1})=Q_{1-}^{+}Q_{2}\sqrt{Q_{3}}$ and $\sqrt{1+(x_{2}+x_{1})^{2}}=Q_{4-}^{+}Q_{5}\sqrt{Q_{3}}$, where each Q is a rational number and $Q_{3}$ must not be a perfect square. Finally, a constructive algorithm for producing N pairwise rational points on the constraint parabola is given. Of course N can be as high as infinity, so the answer to the open problem is affirmative and that an infinite number of families each containing an arbitrarily large number of pairwise rational points on the constraint parabola do exist.

1 Choice of Coordinates and Factoring

To begin, lets use the same x and y coordinates as the constraint parabola $y=x^{2}$ to measure the distance between points. The distance between two arbitrary points on the parabola is then $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ and $y=x^{2}$, $x,y>0$ so the distance between any two points on the parabola is $\sqrt{(x_{2}-x_{1})^{2}+(x_{2}^{2}-x_{1}^{2})^{2}}$. We can factor this expression to obtain $\sqrt{(x_{2}-x_{1})^{2}+(x_{2}+x_{1})^{2}(x_{2}-x_{1})^{2}}$ = $\sqrt{(x_{2}-x_{1})^{2}[1+(x_{2}+x_{1})^{2}]}$ = $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$. Now, there are two possibilities for the distance between the two points $(x_{2}-x_{1})$: it is either rational or irrational. We are only interested in rational distances, so we require that $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$ (the total distance between the two arbitrary points) be rational in both cases.

2 Rational Solutions

The first case in which $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$ is rational (and so the distance between the points $(x_{2}-x_{1})$ is rational) is the situation in which both $(x_{2}-x_{1})$ and $\sqrt{1+(x_{2}+x_{1})^{2}}$ are themselves rational. Let’s look at the circumstances under which $\sqrt{1+(x_{2}+x_{1})^{2}}$ will be rational. Obviously, to meet this condition $1+(x_{2}+x_{1})^{2}$ must be a perfect square. Let $a=x_{2}+x_{1}$ and let the condition become $1+a^{2}$ must be a perfect square. Now we know that a and $a^{2}$ cannot be an integer so a must be of the form $\dfrac{x}{y}$ where y is not equal to 1. In this case we can write $1+a^{2}$ as $1+\dfrac{x^{2}}{y^{2}}$ = $\dfrac{y^{2}}{y^{2}}+\dfrac{x^{2}}{y^{2}}$ = $\dfrac{x^{2}+y^{2}}{y^{2}}$. We know the denominator $y^{2}$ will be a perfect square, so that is taken care of. What we need to ensure is that $x^{2}+y^{2}$ is a perfect square. So we need the solution to $x^{2}+y^{2}=n^{2}$, which of course we know are the Pythagorean triples. So for each Pythagorean triple we get an x, y, and n, and $x_{2}-x_{1}=\dfrac{x}{y}$ is a solution (the distance between the two points along the x axis) to obtain two points that are a rational distance apart. So for two points on the parabola to be a rational distance apart, the sum of their distances from the x axis ($x_{2}+x_{1}$) must be $\dfrac{x}{y}$ or $\dfrac{y}{x}$ where $x^{2}+y^{2}=n^{2}$ or $x^{2}+y^{2}$ is a perfect square or x and y must be the first two terms in a given Pythagorean triple. This type of fraction I have been referring to as a Pythagorean fraction. This family of solutions is perhaps thwarted by my own conjecture that the difference of two Pythagorean fractions cannot be equal for two different sets of Pythagorean fractions. It will perhaps be more fruitful to search for a more general family of solutions.

3 Irrational Solutions

3.1 Q $\sqrt{\dfrac{n}{m}}$ Solution Family

Let’s explore the more general case in which distance between the x coordinates of the points $(x_{2}-x_{1})$ is equal to $\sqrt{1+(x_{2}+x_{1})^{2}}$ first. Such a family of solutions forces $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$ to be rational and allows $(x_{2}-x_{1})$ to be irrational in the form of $\sqrt{\dfrac{n}{m}}$ where n,m=integers, has $(x_{2}-x_{1})$ equal to $\sqrt{1+(x_{2}+x_{1})^{2}}$ times some arbitrary rational number. For example, we can have $(x_{2}-x_{1})$ = $2\sqrt{3}$ (irrational) and $\sqrt{1+(x_{2}+x_{1})^{2}}$ = $\sqrt{3}$ (irrational), and then the total distance $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$ = 2*3=6 (rational). Now this will ensure that the distance between the points is rational if the distance between the x coordinates of the two points $(x_{2}-x_{1})$ is an irrational number of the form $\sqrt{\dfrac{n}{m}}$ where n and m are integers, but this constraint (that $(x_{2}-x_{1})$ be equal to $\sqrt{1+(x_{2}+x_{1})^{2}}$) will of course ensure that the distance between the two points is rational in the case that $(x_{2}-x_{1})$ is rational as well. To enforce these conditions we must set $(x_{2}-x_{1})$ = C$\sqrt{1+(x_{2}+x_{1})^{2}}$, where C is an arbitrary rational number. Rearraging terms we obtain: $(x_{2}-x_{1})^{2}-C^{2}(x_{2}+x_{1})^{2}=C^{2}$. This is the equation of a hyperbola. Any point on the this hyperbola (x,y) where (x - y) is irrational represents a valid set of two points that are a rational distance away from each other and lie on the parabola $y=x^{2}$. It is important to note that this conclusion holds whether $(x_{2}-x_{1}),x_{2},$ and $x_{1}$ are rational or irrational as long as $(x_{2}-x_{1})$ = C$\sqrt{1+(x_{2}+x_{1})^{2}}$. This generates an infinite family of hyperbolas satisfying the equation $(x-y)$ = C$\sqrt{1+(x+y)^{2}}$ indexed by the arbitrary rational number C. Unfortunately, we can see by cursory analysis or by computer graphing software, that these equations admit no solutions where $x,y>0$ and as such are irrelevant to our problem. One way not to invent a lightbulb I suppose.

3.2 $Q_{1}\sqrt{\dfrac{n}{m}}+Q_{2}$ Family of Solutions

Perhaps we should investigate another, more general, family of solutions that force $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$ to be rational. This family consists of solutions where $(x_{2}-x_{1})$ = $Q^{+}_{-}R$ and $\sqrt{1+(x_{2}+x_{1})^{2}}$ = $Q^{-}_{+}R$ where Q is an arbitrary member of the Rationals and R is a member of the Reals of the form $\sqrt{\dfrac{n}{m}}$ with m,n being integers and $\dfrac{n}{m}$ not being equal to a perfect square. If we hold to this constraint, we get the equation $(x_{2}-x_{1})+2R$ = $C\sqrt{1+(x_{2}+x_{1})^{2}}$. Taking 2R inside the parentheses and squaring both sides, we obtain: $(x_{2}-x_{1}+2R)^{2}$ = $C^{2}+C^{2}(x_{2}+x_{1})^{2}$ or in more standard hyperbolic form: $(x_{2}-x_{1}+2R)^{2}-C^{2}(x_{2}+x_{1})^{2}=C^{2}$. This equation does admit solutions for $x,y>0$. In order to generate a subset of this infinite family of solutions, we need to simultaneously satisfy two equations: 1) $(x_{2}-x_{1})$ = $Q^{+}_{-}R$ and 2) $(x_{2}-x_{1}+2R)^{2}$ = $C^{2}+C^{2}(x_{2}+x_{1})^{2}$. This means that the distance between every two points must be of the form $Q^{+}_{-}R$ and each of the two points must be the x and y coordinates of a point that lies on the hyperbola $(x_{2}-x_{1}+2R)^{2}$ = $C^{2}+C^{2}(x_{2}+x_{1})^{2}$.

Both of these graphs are difficult to read and get explicit solutions out of so I will instead seek to satisfy the original conditions of this family of solutions directly: 1) $(x_{2}-x_{1})$ is of the form $Q_{1}+Q_{2}\sqrt{\dfrac{n}{m}}$ where $Q_{1}$ and $Q_{2}$ are rational numbers and n,m are integers with $\dfrac{n}{m}$ not equal to a perfect square. 2) $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$ is rational. To enforce these conditions, we will set $\sqrt{1+(x_{2}+x_{1})^{2}}$ equal to $Q_{1}+Q_{2}\sqrt{\dfrac{n}{m}}$ and $(x_{2}-x_{1})$ = $Q_{1}-Q_{2}\sqrt{\dfrac{n}{m}}$ so that $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$ = $Q_{1}^{2}+Q_{2}^{2}\dfrac{n}{m}$. Upon picking the correct $Q_{1}$, $Q_{2}$, and keeping $\sqrt{\dfrac{n}{m}}$ constant, we can begin the process of constructing what I’ve been calling ”John Diamonds,” after my uncle who passed away earlier this week and who I miss dearly. The procedure roughly goes as follows:

1. Select n and m and calculate $\sqrt{\dfrac{n}{m}}$, such that $\dfrac{n}{m}$ is not a perfect square, which we will name R. Select $Q_{1}$ and $Q_{2}$. Then set $x_{2}-x_{1}$ = $Q_{1}-Q_{2}R$ and $\sqrt{1+(x_{2}+x_{1})^{2}}$ = $Q_{1}+Q_{2}R$. This enforces the condition that $(x_{2}-x_{1})\sqrt{1+(x_{2}+x_{1})^{2}}$ is rational. Now mark the point of intersection between the two equations.

2. To find N more points that are a rational distance away from the point we found in the first step, set $x_{2}-x_{1}$ = $Q_{1}+nQ-Q_{2}R$ and $\sqrt{1+(x_{2}+x_{1})^{2}}$ = $Q_{1}+nQ+Q_{2}R$ (where n is an integer) and find the points of intersection. This should give you N equally spaced points along the horizontal axis with horizontal spacing $\dfrac{N}{n}$ between each point. Repeat this procedure with Q equal to N different rational numbers to create a line of solutions where the two equations intersect all of which have the same y coordinate.

3. Now, for each of the intersection points along the horizontal line we found in step two indexed by n, set the equations equal to $x_{2}-x_{1}$ = $Q_{1}+nQ-(Q_{2}+Q_{n})R$ and $\sqrt{1+(x_{2}+x_{1})^{2}}$ = $Q_{1}+nQ+(Q_{2}+Q_{n})R$, where n is an integer that specifies the location of the points in step two along the horizontal line of points we found in that step, and $Q_{n}$ are a set of n equally spaced rational numbers. The spacing between adjacent $Q_{n}$ and adjacent nQ should be equal. You should now have an NxN grid of intersection points. If we index each intersection point we have found by n as $(x_{n},y_{n})$, then the NxN grid of points should cover every permutation of $(x_{n},y_{n})$ where n runs from 1 through N. This proves that every permutation of the intersection points is itself an intersection point and that they are all pairwise rational distances from each other. One can explicitly construct such an NxN grid of such points and verify themselves that all the points therein are pairwise rational distances from each other.

For example if we choose the initial point to be the intersection of $\sqrt{5}-8$ = $x_{2}-x_{1}$, $\sqrt{5}+8$ = $\sqrt{1+(x_{2}+x_{1})^{2}}$ we can attain a vertical line of n intersection points representing points that are a rational distance away from this initial point by choosing a series of points $n\sqrt{5}-8$ = $x_{2}-x_{1}$ and where n are the natural numbers. $\sqrt{1+(x_{2}+x_{1})^{2}}$ = $n\sqrt{5}+8$. You can then get a grid of John Diamonds by choosing $n\sqrt{5}-(8+m)$ = $x_{2}-x_{1}$ and $n\sqrt{5}+(8+m)$ = $\sqrt{1+(x_{2}+x_{1})^{2}}$ and plot the points of intersection for each n and m, where n and m are natural numbers. To find the x and y coordinates of each intersection points you can use computer graphing software or simply solve the set of equations explicitly for $x_{1}$ and $x_{2}$. Each set of $x_{1}$ and $x_{2}$ are the x coordinates of a set a points on the parabola that are a rational distance away from each other. By constructing John Diamonds, one can find a set of as many points, each of which are pairwise a rational distance away from each other, as one desires.

The graph of John Diamonds for N=3 is shown below:

4 Conclusion

This paper has presented a heretofore, unsolved problem in Number Theory and Geometry of public interest and then proceeded to work out the solution. The problem can be answered affirmatively. There are in fact an infinite number of families of intersection points each indexed by the irrational square root of a rational number that is not a perfect square. Depending on how far from the x=0 axis we placed the original chosen point that generates the family, we could find N points, each indexed by two rational numbers, on the constraint parabola that were a pairwise rational distance away from each other. We can make the chosen origin point an arbitrarily large distance from x=0. This way the number of points on the constraint parabola $y=x^{2}$ that are a pairwise rational distance away from each other can be made as large as desired. The proof method in this paper was chosen purposely to be constructive. In fact, it was simply proof by constructive algorithm. This was done so that any reader could generate as many pairwise rational points on the parabola as she desires using the methods laid out in this paper.

5 Citations

[1] Nate Dean. Geometry/Number Theory Open Problems: Points on a Parabola

http://dimacs.rutgers.edu/ hochberg/undopen/geomnum/geomnum.html